Let $y=x^2\ln(x)$. $\dfrac{dy}{dx}=$
Answer: $x^2\ln(x)$ is the product of two, more basic, expressions: $x^2$ and $\ln(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}(x^2\ln(x)) \\\\ &=\dfrac{d}{dx}(x^2)\ln(x)+x^2\dfrac{d}{dx}(\ln(x))&&\gray{\text{The product rule}} \\\\ &=2x\cdot \ln(x)+x^2\cdot \dfrac1x&&\gray{\text{Differentiate }x^2\text{ and }\ln(x)} \\\\ &=2x\ln(x)+x&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=2x\ln(x)+x$ or any other equivalent form.